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If $X$ is a compact metric space, $A: X\to X$, is it true that if $a = \inf d(x,Ax),\space x \in X$, then there exists $y \in X$ such that $d(y,Ay) = \inf d(x,Ax)$? If so, why?

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No, it's not true. Let $X = [0,1]$. Define $A$ by $Ax = 1$ for $x\in [0,1)$, and $A1 = 0$. Then $a=0$ but always $d(x,Ax) > 0$. (Obviously, the statement is true if $A$ is continuous.)

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oh wait... I get it –  Atreyu Oct 14 '12 at 22:33
can I refer you to this proof... math.stackexchange.com/questions/118536/…. The answer uses the fact that K is compact to say that this holds... –  Atreyu Oct 14 '12 at 22:38
In that question $A$ is a contraction, therefore, it is a continuous function. –  Yury Oct 14 '12 at 22:43
if A is a contraction, wouldn't it follow immediately that there exists a unique fixed point –  Atreyu Oct 14 '12 at 22:45
It depends on your definition of “immediately” :-). Yes, it is a true that every contraction $A$ has a unique fixed point, if $X$ is a compact metric space (or if $X$ is a complete metric space). –  Yury Oct 14 '12 at 22:56

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